Problem: Is ${798181}$ divisible by $3$ ?
Explanation: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {798181}= &&{7}\cdot100000+ \\&&{9}\cdot10000+ \\&&{8}\cdot1000+ \\&&{1}\cdot100+ \\&&{8}\cdot10+ \\&&{1}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {798181}= &&{7}(99999+1)+ \\&&{9}(9999+1)+ \\&&{8}(999+1)+ \\&&{1}(99+1)+ \\&&{8}(9+1)+ \\&&{1} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {798181}= &&\gray{7\cdot99999}+ \\&&\gray{9\cdot9999}+ \\&&\gray{8\cdot999}+ \\&&\gray{1\cdot99}+ \\&&\gray{8\cdot9}+ \\&& {7}+{9}+{8}+{1}+{8}+{1} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${798181}$ is divisible by $3$ if ${ 7}+{9}+{8}+{1}+{8}+{1}$ is divisible by $3$ Add the digits of ${798181}$ $ {7}+{9}+{8}+{1}+{8}+{1} = {34} $ If ${34}$ is divisible by $3$ , then ${798181}$ must also be divisible by $3$ Add the digits of ${34}$ $ {3}+{4} = \color{#9D38BD}{7} $ If $\color{#9D38BD}{7}$ is divisible by $3$ , then ${34}$ must also be divisible by $3$ $\color{#9D38BD}{7}$ is not divisible by $3$, therefore ${798181}$ must not be divisible by $3$.